In programming contests, your program needs to finish running within a certain timeframe in order to receive credit. For USACO, this limit is $2$ seconds for C++ submissions, and $4$ seconds for Java/Python submissions. A conservative estimate for the number of <TextTooltip content={`You can think of an "operation" as one instruction for the computer. For example, multiplying two numbers together, reading in a number from the input, or outputting "Hello World" would all be considered "operations." `}>operations</TextTooltip> the grading server can handle per second is $10^8$, but it could be closer to $5 \cdot 10^8$ given good constant factors<Asterisk>If you don't know what constant factors are, don't worry -- we'll explain them below.</Asterisk>.
We want a method to calculate how many operations it takes to run each algorithm, in terms of the input size $n$. Fortunately, this can be done relatively easily using [Big O Notation](https://en.wikipedia.org/wiki/Big_O_notation), which expresses worst-case time complexity as a function of $n$ as $n$ gets arbitrarily large. Complexity is an upper bound for the number of steps an algorithm requires as a function of the input size. In Big O notation, we denote the complexity of a function as $O(f(n))$, where constant factors and lower-order terms are generally omitted from $f(n)$. We'll see some examples of how this works, as follows.
We can find the time complexity of multiple loops by multiplying together the time complexities of each loop. This example is $O(nm)$, because the outer loop runs $O(n)$ iterations and the inner loop $O(m)$.
In this example, the outer loop runs $O(n)$ iterations, and the inner loop runs anywhere between $1$ and $n$ iterations (which is a maximum of $n$). Since Big O notation calculates worst-case time complexity, we treat the inner loop as a factor of $n$.<Asterisk>We can also do some math to calculate exactly how many times the code runs: 1+2+...+n = n*(n-1)/2 = (n^2 - n)/2 = O(n^2)</Asterisk> Thus, this code is $O(n^2)$.
If an algorithm contains multiple blocks, then its time complexity is the worst time complexity out of any block. For example, the following code is $O(n^2)$.
The following code is $O(n^2 + m)$, because it consists of two blocks of complexity $O(n^2)$ and $O(m)$, and neither of them is a lower order function with respect to the other.
The "Constant Factor" of an algorithm refers to the coefficient of the complexity of an algorithm. If an algorithm runs in $O(kn)$ time, where $k$ is a constant and $n$ is the input size, then the "constant factor" would be $k$.
Normally when using the big-O notation, we ignore the constant factor: $O(3n) = O(n)$. This is fine most of the time, but sometimes we have an algorithm that just barely gets TLE, perhaps by just a few hundred milliseconds. When this happens, it is worth optimizing the constant factor of our algorithm. For example, if our code currently runs in $O(n^2)$ time, perhaps we can modify our code to make it run in $O(n^2/32)$ by using a bitset. (Of course, with big-O notation, $O(n^2) = O(n^2/32)$.)
For now, don't worry about how to optimize constant factors -- just be aware of them.
Here are conservative upper bounds on the value of $n$ for each time complexity. You can probably get away with more than this, but this should allow you to quickly check whether an algorithm is viable.